Sets: Class 11 Maths Chapter 1 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

105 min
56 practice questions
26 MCQs

Is chapter mein aap kya seekhenge

Decide karo ki koi collection well defined hai ya nahi, aur set ko roster form aur set-builder form mein describe karo.
Empty, singleton, finite, infinite aur equal sets pehchano, aur finite set ke distinct elements gino.
Membership (∈) aur inclusion (⊂) mein farak samjho, chhote set ke saare subsets ki list banao aur proper subsets pehchano.
R ke standard subsets ke naam batao aur interval notation aur set-builder form ke beech convert karo.
Sahi universal set chuno aur sets aur unke beech ke relations ke liye Venn diagrams banao.
Unions, intersections, differences aur complements nikalo, aur unke laws lagao, distributive aur De Morgan laws samet.
Set identities prove karo yeh dikhakar ki har side doosri side ka subset hai, aur galat claims ko counterexample se disprove karo.

NCERT Mathematics Class 11, Chapter 1, pp. 1–23 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

1.1 Introduction

Squad chunta hua selector, ek railway app jo sirf Bhopal se 6 pm ke baad chalne wali trains dikhata hai, ek phone jo contacts ko "Family" aur "School" mein rakhta hai: har ek pehle ek collection banata hai aur phir ek-ek object ke liye tay karta hai ki kaun andar hai aur kaun bahar. Mathematics roz ki is aadat ko ek precise tool mein badal deta hai jise set (समुच्चय) kehte hain.

Sets is book ke baaki hisse ki common language hain: function (Chapter 2) ordered pairs ka ek set hai, inequality (Chapter 5) ke solutions number line par ek set banate hain, aur probability (Chapter 14) mein event outcomes ka ek set hai. Yeh chapter sets ko describe, compare aur combine karne ki vocabulary banata hai. Yeh theory 1870s mein Georg Cantor (1845–1918) ne shuru ki; unki kahani historical note mein hai.

1.2 Sets aur unke Representations

Hum bina soche hamesha cheezon ko group karte rehte hain: playlist ke songs, local train ke passengers, tumhari desk par rakhi books. Lekin ek mathematical collection ko ek sakht test paas karna padta hai. Koi bhi object uthao, aur sawaal "kya yeh belong karta hai?" ka jawaab pakka haan ya nahi hona chahiye.

Kuch descriptions ko test karo. "India ke woh states jinki border Nepal se milti hai" pass hota hai: Uttarakhand, Uttar Pradesh, Bihar, West Bengal aur Sikkim belong karte hain, aur Kerala nahi. " $x^{2} - 7 x + 12 = 0$ ke roots" bhi pass hota hai, jismein theek 3 aur 4 belong karte hain. "India ke sabse mahaan spinners" fail hota hai, kyunki do cricket fans imaandaari se bhi alag raay rakh sakte hain. "Tumhari class ke lambe students" bhi isi wajah se fail hota hai, lekin ise theek karna aasaan hai: "tumhari class ke 165 cm se lambe students" ek bilkul sahi set hai.

Definition

Set

Set objects ka aisa collection hai jo well defined ho: har object ke liye saaf pata ho ki woh collection mein belong karta hai ya nahi. In objects ko set ke elements ya members kehte hain.

Example20 se chhote prime numbers ek set banate hain; 20 se chhote "interesting" numbers nahi.

Hum sets ke naam capital letters jaise $A$ , $B$ , $S$ , $X$ se rakhte hain aur elements ke naam small letters jaise $a$ , $p$ , $x$ se. Symbol $\in$ (Greek epsilon ka ek stylised roop) ka matlab hai "ka element hai" ya "belong karta hai", aur $\notin$ ka matlab hai "ka element nahi hai". Agar $P$ 20 se chhote primes ka set hai, to $7 \in P$ lekin $9 \notin P$ , kyunki $9 = 3 \times 3$ . Agar $M$ un months ka set hai jinmein theek 30 din hote hain, to June $\in M$ aur July $\notin M$ .

Table 1.1Is poori book mein use hone wale number setsNCERT 1.2 · p. 2

Symbol	Kiska set	Kuch members	Member nahi
$𝐍$	natural numbers	$1, 2, 3, 100$	$0$
$𝐙$	integers	$- 3, 0, 5$	$\frac{1}{2}$
$𝐐$	rational numbers	$- \frac{2}{3}, 0.25, 7$	$\sqrt{2}$
$𝐑$	real numbers	$\sqrt{2}, π, - 4.1$	$\sqrt{- 1}$
$𝐙^{+}$	positive integers	$1, 2, 3$	$0$
$𝐐^{+}$	positive rationals	$\frac{3}{4}, 5$	$- \frac{3}{4}$
$𝐑^{+}$	positive reals	$π, 0.001$	$0$

Is book mein natural numbers 1 se shuru hote hain, isliye $0 \notin 𝐍$ aur $𝐙^{+}$ wahi set hai jo $𝐍$ hai. $𝐑$ ke bahar ke numbers, jaise $\sqrt{- 1}$ , tum Chapter 4 mein dekhoge.

Set ko describe karne ke do standard tareeke hain. Roster form mein hum elements ko ek-ek karke curly braces ke beech likhte hain, beech mein commas lagakar: 28 ke positive divisors se yeh set banta hai: ${1, 2, 4, 7, 14, 28} .$ Teen conventions notation ko saaf rakhte hain.

Order se farak nahi padta. ${1, 2, 4, 7, 14, 28}$ aur ${28, 7, 1, 14, 4, 2}$ ek hi set ko describe karte hain.
Har element ek hi baar likha jaata hai. Word COMMITTEE ke letters ka set nau letters ki list nahi hai, balki ${C, O, M, I, T, E},$ jismein har letter ek baar hai.
Teen dots ek saaf pattern ko aage badhate hain. 5 ke positive multiples hain ${5, 10, 15, 20, \dots},$ aur dots ka matlab hai ki pattern kabhi rukta nahi.

Set-builder form (समुच्चय निर्माण रूप) kuch list nahi karta. Iske badle yeh ek test deta hai: ek condition jo set ke members paas karte hain aur baaki har object fail karta hai. 28 ke divisors ban jaate hain ${x : x \in 𝐍 and x divides 28};$ shabdon mein, "har natural number $x$ jo 28 ko divide karta hai". Letter $x$ sirf ek placeholder hai jiski jagah koi bhi doosra letter aa sakta hai, aur kuch books colon ki jagah vertical bar use karti hain, jaise ${x ∣ x > 0}$ .

3/4Ek rule se describe kiya gaya jise yeh chhahon numbers satisfy karte hain aur koi doosra number nahi karta.

Fig. 1.1Ek hi set: shabdon mein, list ke roop mein aur rule ke roop mein.

Chhah number tiles, 1, 2, 4, 7, 14 aur 28, heading "28 ke positive divisors" ke neeche. Pehle step mein tiles bikhri hui hain, ek collection jo sirf shabdon mein describe hai. Doosre step mein woh do bade curly braces ke beech ek row mein lag jaati hain, jo roster form hai. Teesre step mein row wahi rehti hai aur uske neeche set-builder rule aata hai: x aisa ki x ek natural number hai jo 28 ko divide karta hai. Chauthe step mein tiles jagah badal kar order 14, 1, 28, 2, 7, 4 mein aa jaati hain aur pehle 7 ke neeche 7 ki ek dashed doosri copy note "ek hi baar" ke saath dikhti hai: set badla nahi hai.

Har form ka apna kaam hai. Roster form chhote sets ke liye theek hai aur ek nazar mein membership check karne deta hai. Set-builder form un sets ko sambhalta hai jo bade, infinite ya list na ho sakne wale hain, jaise ${x : x \in 𝐑, 0 < x < 1},$ yaani 0 aur 1 ke beech ke real numbers. Dono forms ke beech aana-jaana ek basic skill hai, aur neeche ke examples dono directions ki practice karwate hain.

Example 1NCERT Example 1 · p. 3

Quadratic ke roots ek set ke roop mein

$2 x^{2} - 5 x - 3 = 0$ ka solution set roster form mein likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2NCERT Example 2 · p. 3

Rule se list tak

${n : n \in 𝐙, - 2 \leq 3 n + 1 < 11}$ ko roster form mein likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3NCERT Example 3 · p. 4

Rule pehchanna

$A = {2, 5, 10, 17, 26, \dots}$ ko set-builder form mein likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4NCERT Example 4 · p. 4

Fractions mein ek pattern

${\frac{3}{4}, \frac{5}{9}, \frac{7}{16}, \frac{9}{25}, \frac{11}{36}}$ ko set-builder form mein likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5NCERT Example 5 · p. 4

Do descriptions ko match karna

Roster form ke har set ko set-builder form mein likhe usi set se match karo.

Roster form	Set-builder form
(i) ${K, O, L, A, T}$	(a) ${x : x \in 𝐙, x^{2} = 4}$
(ii) ${1}$	(b) word KOLKATA ke letters
(iii) ${1, 3, 9}$	(c) ${x : x \in 𝐍, x divides 9}$
(iv) ${- 2, 2}$	(d) ${x : x \in 𝐑, x^{3} = 1}$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1.3 Empty Set

Kuch bilkul saaf descriptions kuch bhi select nahi karte. Jaise ${x : x \in 𝐍 and x + 5 = 2} :$ $x + 5 = 2$ wala akela number $- 3$ hai, jo natural nahi hai, isliye koi element qualify nahi karta. Ya phir apne school ke Class 11 ke woh students lo jo 30 February ko paida hue. Description precise hai, aur answer hai "koi nahi".

Definition

Empty set

Jis set mein ek bhi element na ho, use empty set kehte hain; iske liye null set aur void set naam bhi use hote hain. Hum ise $ϕ$ ya ${}$ likhte hain.

Example ${x : x \in 𝐑, x^{2} + 1 = 0} = ϕ,$ kyunki kisi real number ka square kabhi $- 1$ nahi hota.

Teen aur empty sets, har ek apne reason ke saath:

${x : x \in 𝐍, x < 1} = ϕ,$ kyunki sabse chhota natural number 1 hai.
${x : x \in 𝐐, x^{2} = 3} = ϕ,$ kyunki iske sirf do solutions, $\pm \sqrt{3}$ , irrational hain.
Jo primes 4 ke multiples hain, unka set $ϕ$ hai: 4 ka har multiple 2 se divisible hai aur 2 se bada hai, isliye woh prime nahi hai.

Braces ke saath dhyaan rakho. $ϕ$ ek khaali box hai, ${0}$ ek box hai jismein number 0 hai, aur ${ϕ}$ ek box hai jismein ek khaali box hai. Sirf pehla wala empty hai (Fig. 1.2).

3/3 ${ϕ}$ mein ek element hai, khud empty set.

Fig. 1.2Sirf

ϕ

empty hai;

{0}

aur

{ϕ}

mein ek-ek element hai.

Ek row mein teen rounded boxes. Pehle step mein sirf left box dikhta hai: woh khaali hai, uska label hai phi barabar khaali braces, aur note "no element". Doosre step mein beech ka box aata hai jismein bada 0 hai, label 0 wala set, note "ek element" ke saath. Teesre step mein right box aata hai jismein ek chhota khaali box hai, label phi wala set, aur iska note bhi "ek element" hai.

Quick check

2 mein se 1

Inmein se kaunse sets empty hain?

{x : x \in 𝐍, 2 x = 7}

{x : x \in 𝐙, x^{2} = 0}

{x : x \in 𝐑, x^{2} < 0}

{ϕ}

Saare sahi options chuno

Answer dekho

Answer

(A) ${x : x \in 𝐍, 2 x = 7}$
(C) ${x : x \in 𝐑, x^{2} < 0}$

2 x = 7

ke liye

x = 3.5

chahiye, jo natural nahi hai, aur koi real square negative nahi hota. Lekin

x^{2} = 0

ka solution 0 hai, aur

{ϕ}

mein ek element hai, empty set.

1.4 Finite aur Infinite Sets

Ek set kitna bada hai? ${a, e, i, o, u}$ ke liye answer 5 hai. Set $A$ ke distinct elements ki sankhya ko hum $n (A)$ likhte hain, isliye dohraayi gayi entry kabhi do baar nahi gini jaati. Agar $L$ , BANANA ke letters ka set hai, to $L = {B, A, N}$ aur $n (L) = 3.$

Kuch sets bahut bade hote hain, phir bhi principle mein gine ja sakte hain. Kerala mein 14 districts hain, aur Juhu beach ki ret ke daane itne hain ki soch bhi nahi sakte, phir bhi yeh ginti kahin na kahin ruk jaati hai. Kuch sets kabhi khatam nahi hote: 7 ke multiples ko tum kitna bhi aage tak list karo, hamesha ek agla multiple hota hai (Fig. 1.3).

Definition

Finite aur infinite sets

Set finite hota hai agar woh empty ho ya uske elements ki ginti kisi natural number $n$ par khatam ho jaaye; tab $n (A) = n,$ aur $n (ϕ) = 0.$ Jo set finite nahi hai, woh infinite hai.

ExampleHafte ke din $n = 7$ wala finite set banate hain; 7 ke positive multiples ek infinite set banate hain.

3/3Din $n = 7$ par ruk jaate hain, yeh finite set hai. 7 ke multiples ka koi last element nahi hai, yeh infinite set hai.

Fig. 1.3Finite set ke naye elements khatam ho jaate hain; infinite set ke kabhi nahi hote.

Tiles ki do rows. Pehle step mein har row mein teen tiles dikhti hain: upar wali row mein Mon, Tue, Wed aur neeche wali row mein 7, 14, 21. Doosre step mein dono rows bhar jaati hain: upar wali row saat tiles ke baad Sun tak pahunchti hai, aur neeche wali row 49 tak chalti hai. Teesre step mein upar wali row ko label n = 7 milta hai, jabki neeche wali row teen dots aur kinaare se bahar jaate ek arrow par khatam hoti hai, jiska label hai "no last element".

Aur examples: Kerala ke districts $n = 14$ wala finite set banate hain; $x^{3} = x$ ke solutions ka set hai ${- 1, 0, 1},$ jo $n = 3$ wala finite set hai; circle par ke points aur kisi diye gaye point se guzarne wali lines infinite sets banati hain.

Roster form infinite set ko poora list nahi kar sakta, isliye hum itne elements likhte hain ki pattern bilkul saaf ho jaaye aur phir dots lagate hain: 7 ke positive multiples ke liye ${7, 14, 21, \dots}$ , ya integers mein 3 ke multiples ke liye ${\dots, - 6, - 3, 0, 3, 6, \dots}$ . Kuch infinite sets ke saath yeh bhi nahi chalta. 0 aur 1 ke beech ke real numbers kisi aise pattern ko follow nahi karte jise dots aage badha sakein, aur Cantor ne prove kiya ki koi bhi list un sabko contain nahi kar sakti (historical note dekho). Aise sets ke liye set-builder form ya interval notation (Section 1.6.2) hi ek tareeka hai.

Example 6NCERT Example 6 · p. 7

Finite ya infinite?

Decide karo ki har set finite hai ya infinite.

${x : x \in 𝐍, (x - 2) (x + 5) = 0}$
${x : x \in 𝐍, x^{2} = 5}$
${x : x \in 𝐙, ∣ x ∣ < 4}$
${x : x \in 𝐍, x is a multiple of 3}$
${x : x \in 𝐑, 0 < x < 0.001}$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Kaunsa set infinite hai?

point

(1, 2)

se guzarne wali linesaaj India mein rehne wale log

x^{4} = 16

ke real rootsEnglish alphabet ke letters

Answer dekho

Answer (A) point $(1, 2)$ se guzarne wali lines

Ek point se har direction mein ek line jaati hai, isliye aisi infinitely many lines hain. India ki population bahut badi hai lekin finite hai, aur

x^{4} = 16

ke sirf real roots

\pm 2

hain.

1.5 Equal Sets

Do descriptions kab ek hi set dete hain? Sirf elements maayne rakhte hain: na wording, na order, na repeats.

Definition

Equal sets

Do sets equal hote hain, $A = B$ , jab ek mein belong karne ka matlab hamesha doosre mein belong karna ho: $A$ ka har element $B$ mein ho aur $B$ ka har element $A$ mein ho. Agar ek bhi element is baat ko todta hai, to $A \neq B$ .

Example ${x : x \in 𝐙, x^{2} = 4}$ aur ${2, - 2}$ equal hain, halaanki ek equation se diya gaya hai aur doosra list se.

Equality membership ke alawa sab kuch ignore karti hai. NAINITAL ke letters aur LATIN ke letters ek hi set dete hain, ${N, A, I, T, L},$ halaanki ek word mein aath letters hain aur doosre mein paanch (Fig. 1.4). Kisi element ko doosri baar list karne se kuch naya nahi judta.

3/3Ek set ka har letter doosre mein hai aur doosre ka har letter pehle mein, isliye sets equal hain.

Fig. 1.4Alag words, letters ka same set.

Letter tiles ki do rows: upar N, A, I, N, I, T, A, L aur neeche L, A, T, I, N. Pehle step mein saari tiles dikhti hain. Doosre step mein NAINITAL ka doosra N, doosra I aur doosra A dhundhle ho jaate hain, aur paanch distinct letters bachte hain. Teesre step mein ek line upar ke har bache letter ko neeche wali row ke usi letter se jodti hai: har letter ka theek ek partner hai, isliye dono sets equal hain.

$A \neq B$ dikhane ke liye ek witness kaafi hai: aisa element jo ek set mein ho aur doosre mein na ho. Equal sets mein hamesha elements ki sankhya same hoti hai, lekin converse sahi nahi hai: ${1, 2}$ aur ${3, 4}$ dono mein do elements hain aur koi bhi common nahi.

Example 7NCERT Example 7 · p. 8

Equal sets chhaantna

Inmein se kaunse sets equal hain?

$P = {- 2, 2}$
$Q = {x : x \in 𝐙, x^{2} = 4}$
$R = {x : x \in 𝐑, x^{3} = 8}$
$S$ = even primes ka set
$T = {x : x \in 𝐙, ∣ x ∣ \leq 2}$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8NCERT Example 8 · p. 8

Equal ya nahi?

Har pair ke sets equal hain ya nahi, decide karo aur apne answer ka reason do.

$X$ , yaani TEEN ke letters ka set, aur $Y$ , yaani NET ke letters ka set.
$A$ , yaani 45 ke prime factors ka set, aur $B = {x : x \in 𝐍, x^{2} - 8 x + 15 = 0} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Agar

n (A) = n (B),

A = B

SahiGalat

Answer dekho

Answer Galat

{1, 2}

aur

{3, 4}

mein do-do elements hain lekin woh equal nahi hain. Ulta statement sahi hai: equal sets mein elements ki sankhya same hoti hai.

1.6 Subsets

Har square ek rectangle hai, lekin har rectangle square nahi hota. 2 se bada har prime odd hai, lekin 9 odd hai aur prime nahi. Aise statements, jinmein ek set poori tarah doosre set ke andar hota hai, subset ke idea se samjhe jaate hain.

Definition

Subset

Set $A$ , set $B$ ka subset hai, jise $A \subset B$ likhte hain, agar $A$ ka har member $B$ mein bhi belong karta ho. Agar $A$ ka kam se kam ek member $B$ ke bahar ho, to $A$ , $B$ ka subset nahi hai, jise $A \subset̸ B$ likhte hain.

Example ${2, 4} \subset {1, 2, 3, 4},$ lekin ${2, 5} \subset̸ {1, 2, 3, 4},$ kyunki 5 nahi hai.

Arrow $\Rightarrow$ ka matlab hai "implies", isliye definition ko chhote mein likh sakte hain: $A \subset B$ agar $x \in A \Rightarrow x \in B .$ Is asymmetry par dhyaan do. $A \subset B$ prove karne ke liye tumhe $A$ ke har element ko check karna padega, lekin $A \subset̸ B$ prove karne ke liye $B$ ke bahar $A$ ka ek hi element kaafi hai.

Dono directions mein inclusion ho to equality honi hi hai: agar $A \subset B$ aur $B \subset A$ , to kisi bhi set mein aisa element nahi ho sakta jo doosre mein na ho. Ulta, equal sets ek doosre ke subsets hote hain. Aise two-way statement ke liye double arrow $\Leftrightarrow$ use hota hai, jise "if and only if" padhte hain:

\begin{array}{c} A = B \\ ⇕ \\ A \subset B aur B \subset A \end{array}

Yeh two-way check do sets ko equal prove karne ka standard tareeka hai, aur tum ise Miscellaneous Examples mein use karoge.

Definition se seedhe do facts nikalte hain. Har set khud ka subset hai, $A \subset A$ , kyunki $A$ ka har element $A$ mein hai. Empty set har set ka subset hai, $ϕ \subset A$ : rule todne ke liye $ϕ$ ke paas $A$ ke bahar ka ek element hona chahiye, aur uske paas to koi element hi nahi hai.

Jab $A \subset B$ lekin $A \neq B$ , to set $B$ mein kam se kam ek aisa element hai jo $A$ mein nahi hai. Tab $A$ , $B$ ka proper subset (उचित उपसमुच्चय) hai, aur $B$ , $A$ ka superset hai. Jis set mein theek ek element ho, jaise ${7}$ , use singleton kehte hain.

Kuch examples. $𝐙 \subset 𝐐$ , kyunki har integer $m$ barabar hai $\frac{m}{1}$ ke. Squares rectangles ka proper subset banate hain, aur ${4}$ , ${4, 9}$ ka proper subset hai. ${1, 2}$ aur ${2, 3}$ mein koi bhi set doosre ka subset nahi hai.

Example 9NCERT Example 9 · p. 10

Andar hai ya nahi?

Maano $P = {2, 6}$ , $Q = {2, 4, 8}$ aur $R = {2, 4, 6, 8, 10} .$ Har pair ke sets ke beech $\subset$ ya $\subset̸$ likho: (i) $Q, R$ (ii) $R, Q$ (iii) $ϕ, P$ (iv) $P, Q$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10NCERT Example 10 · p. 10

Do sets, koi bhi doosre ke andar nahi

Maano $A$ , 4 ke natural-number multiples ka set hai aur $B$ , 6 ke natural-number multiples ka set. Kya $A \subset B$ ? Kya $B \subset A$ ?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Symbols $\in$ aur $\subset$ mein confusion aasaani se hota hai, aur yeh farak tab bahut maayne rakhta hai jab sets ke elements khud sets hon. Maano $A = {2, {2, 3}, 5} .$ Is set ke teen elements hain: number 2, set ${2, 3}$ aur number 5 (Fig. 1.5). To ${2, 3} \in A$ . Lekin ${2, 3} \subset̸ A,$ kyunki uska element 3, $A$ ke teen elements mein se ek nahi hai: 3 andar wale set ke andar baitha hai, khud $A$ mein nahi. Doosri taraf ${2, 5} \subset A$ , aur ${{2, 3}}$ bhi, yaani woh set jiska akela element ${2, 3}$ hai.

3/4Kya ${2, 3}$ ka har element $A$ ka element hai? 2 hai, lekin 3 nahi, isliye ${2, 3} \subset̸ A .$

Fig. 1.5

A

ka element hona aur

A

ka subset hona alag-alag tests hain.

Ek bada rounded bag jiska label A hai, usmein teen cheezein hain: ek khula 2, ek chhota andar wala bag jismein 2 aur 3 hain, aur ek khula 5. Pehle step mein bag dikhaya gaya hai aur uske neeche set notation mein A likha hai. Doosre step mein andar wala bag highlight hota hai: 2 aur 3 ka set A ka ek element hai. Teesre step mein right side ki do tiles test karti hain ki kya woh set A ka subset hai: tile 2 par "in A" mark hai aur khule 2 ke chaaron taraf ring hai, lekin tile 3 par "not in A" mark hai, kyunki 3 sirf andar wale bag mein hai. Chauthe step mein tiles 2 aur 5 dono par "in A" mark hai aur dono khule numbers par ring hai, isliye 2 aur 5 ka set A ka subset hai.

Example 11NCERT Example 11 · p. 10

Membership aage pahunchti hai; inclusion zaroori nahi

Maano $A \in B$ aur $B \subset C$ . (a) Dikhao ki $A \in C$ . (b) Ek example se dikhao ki $A \subset C$ hona zaroori nahi.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1.6.1 Real numbers ke set ke subsets

Jo number systems tum jaante ho, woh ek doosre ke andar fit hote hain. Set notation mein:

$𝐍 = {1, 2, 3, \dots},$ yaani natural numbers;
$𝐙 = {\dots, - 2, - 1, 0, 1, 2, \dots},$ yaani integers;
$𝐐 = {x : x = \frac{p}{q}, p, q \in 𝐙, q \neq 0},$ yaani rational numbers;
$𝐓 = {x : x \in 𝐑, x \notin 𝐐},$ yaani irrational numbers.

Rational numbers kai roop mein aate hain: $0.75 = \frac{3}{4},$ $0.333 \dots = \frac{1}{3},$ $\sqrt{9} = \frac{3}{1}$ aur $- 1.2 = - \frac{6}{5}$ sab $𝐐$ mein belong karte hain. Jo real number rational nahi hai, woh irrational hai; $\sqrt{3}$ , $\sqrt{7}$ , $π$ aur $0.1010010001 \dots$ iske examples hain, aur inke decimals kabhi kisi repeating block mein nahi bandhte. Basic relations hain $𝐍 \subset 𝐙 \subset 𝐐 \subset 𝐑,$ $𝐓 \subset 𝐑$ aur $𝐍 \subset̸ 𝐓$ (Fig. 1.6).

4/4 $𝐑$ hai $𝐐$ ke saath irrationals $𝐓$ , jaise $\sqrt{2}$ aur $π$ .

Fig. 1.6Har number system apne pehle wale ko contain karta hai;

𝐐

aur

𝐓

milkar

𝐑

banate hain.

Ek ke andar ek rounded rectangles. Pehle step mein sirf sabse andar wala box N dikhta hai, jismein 1, 2 aur 9 hain. Doosre step mein uske chaaron taraf ek bada box Z aata hai aur 0 aur minus 5 jodta hai. Teesre step mein usse bhi bada box Q, Z ko gherta hai aur minus 2.5 aur teen-chauthai jodta hai. Chauthe step mein bahar wala box R aata hai, aur saath mein Q ke bagal mein ek dashed box T, jismein root 2, root 7 aur pi hain; boxes Q aur T overlap nahi karte, aur dono milkar R ko bharte hain.

1.6.2 R ke subsets ke roop mein intervals

$𝐑$ ke kai useful subsets number line ke bina toote hue hisse hote hain, aur unka ek compact notation hai. Maano $a$ aur $b$ real numbers hain, jahan $a < b$ .

Definition

Intervals

Open interval $(a, b) = {x : x \in 𝐑, a < x < b}$ dono end points ko chhod deta hai. Closed interval $[a, b] = {x : x \in 𝐑, a \leq x \leq b}$ dono ko rakhta hai. Half-open intervals $[a, b) = {x : a \leq x < b}$ aur $(a, b] = {x : a < x \leq b}$ theek ek end point rakhte hain.

Example $[2, 5)$ mein 2 aur 4.99 hain lekin 5 nahi.

Square bracket ka matlab hai "included" aur round bracket ka "excluded". Number line par hum included end point ke liye bhara hua dot aur excluded end point ke liye khokhla dot banate hain (Fig. 1.7). In sab intervals mein infinitely many real numbers hote hain, aur number $b - a$ ko uski length kehte hain, brackets chahe jo bhi hon.

3/4Half-open: $- 1 \leq x < 3;$ bhara end belong karta hai, khokhla end nahi.

Fig. 1.7Bhara dot: included. Khokhla dot: excluded.

Minus 3 se 5 tak mark ki gayi number line, jismein minus 1 se 3 tak ek mota highlighted segment hai. Pehle step mein dono end dots khokhle hain: minus 1 se 3 tak ka open interval. Doosre step mein dono dots bhare hain: closed interval. Teesre step mein left dot bhara hai aur right dot khokhla: woh interval jismein minus 1 hai lekin 3 nahi. In teeno steps mein line ke neeche length 4 likhi hai. Chauthe step mein right dot gaayab ho jaata hai aur highlight arrowhead ke saath 5 ke aage tak chala jaata hai: minus 1 se aage ke saare numbers ki ray, jismein minus 1 shaamil hai.

Intervals ek direction mein hamesha ke liye bhi chal sakte hain: $(2, \infty)$ , 2 se bade reals ka set hai, $(- \infty, 5]$ un reals ka set jo zyada se zyada 5 hain, aur $(- \infty, \infty)$ poora $𝐑$ hai. Symbol $\infty$ real number nahi hai, isliye uske saath hamesha round bracket lagta hai; $[2, \infty]$ ka koi matlab nahi hai.

Brackets se subset wale questions jaldi ho jaate hain. Jaise, $(- 1, 4] \subset [- 2, 6),$ kyunki jo bhi $x$ , $- 1 < x \leq 4$ ko satisfy karta hai, woh yeh bhi satisfy karta hai: $- 2 \leq x < 6.$ Notations ke beech convert karna bhi utna hi aasaan hai: ${x : x \in 𝐑, - 2 \leq x < 3} = [- 2, 3),$ aur $(0, 7] = {x : x \in 𝐑, 0 < x \leq 7} .$

Example

Do conditions se ek interval

${x : x \in 𝐑, x \geq - 1 and 2 x - 3 < 5}$ ko interval ke roop mein likho, aur uski length nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1.7 Universal Set

Har discussion ka ek background hota hai. Quadratic equations wale chapter mein tum jo bhi solution set likhte ho, woh $𝐑$ ke andar rehta hai; kisi school ke Class 11 students kaise aate-jaate hain, is survey mein har kaam ka set un students ke set ke andar hota hai.

Definition

Universal set

Universal set, jise $U$ likhte hain, aisa set hai jo kisi diye gaye context mein liye ja rahe saare sets ko contain karta hai, taaki unmein se har ek $U$ ka subset ho.

Example100 se chhote 7 ke multiples ki study ke liye $U$ , ${1, 2, \dots, 99}$ ho sakta hai ya poora $𝐍$ .

$U$ context ke hisaab se kiya gaya ek choice hai, aur yeh choice answers badal sakta hai: $U = 𝐑$ ho to set ${x : x^{2} + 1 = 0}$ empty hai, lekin Chapter 4 number system ko itna bada karta hai ki is equation ke do solutions ho jaate hain.

1.8 Venn Diagrams

Sets ke beech ke relations padhne se zyada dekhne mein aasaan hain. 1880 mein English logician John Venn (1834–1923) ne inke liye ek simple picture di, jise ab Venn diagram (वेन आरेख) kehte hain. Poora rectangle universal set hai; har subset uske andar ek circle ke roop mein banaya jaata hai (koi bhi closed loop chalega), aur elements ko unhi regions mein likha ja sakta hai jahan woh belong karte hain.

Doosre circle ke poori tarah andar bana circle subset dikhata hai, overlap karte circles aise sets dikhate hain jinke kuch elements common hain, aur alag-alag bane circles aise sets dikhate hain jinmein kuch bhi common nahi. Fig. 1.8 mein universal set hai $U = {1, 2, \dots, 12},$ $A$ , $U$ mein 3 ke multiples ka set hai aur $B$ , 6 ke multiples ka set. 6 ka har multiple 3 ka bhi multiple hai, isliye $B$ ka circle $A$ ke circle ke andar hai: $B \subset A$ .

3/3 $B = {6, 12}$ , $A$ ke andar hai, jo dikhata hai ki $B \subset A$ .

Fig. 1.8Circle ke andar circle subset dikhata hai.

U label wale rectangle mein 1 se 12 tak ke numbers hain. Pehle step mein koi circle nahi hai, sirf baarah numbers hain. Doosre step mein 3, 6, 9 aur 12, yaani 3 ke multiples, ke chaaron taraf A label wala circle banta hai. Teesre step mein A ke andar 6 aur 12, yaani 6 ke multiples, ke chaaron taraf B label wala chhota circle aata hai. Baaki aath numbers dono circles ke bahar rehte hain.

Agle section mein tum aur bhi bahut saare Venn diagrams banaoge, jahan sets par har operation ek shaded region ban jaata hai.

Quick check

2 mein se 1

Ek Venn diagram mein

P

ka circle poori tarah

Q

ke circle ke andar hai. Isse kya pata chalta hai?

Q \subset P

P \subset Q

P \in Q

P

aur

Q

mein koi common element nahi hai

Answer dekho

Answer (B) $P \subset Q$

P

ka har element

Q

ke andar bana hai, aur yahi to

P \subset Q

hai. Ek circle ko doosre ke andar banana inclusion dikhata hai, membership nahi:

P \in Q

ka matlab hota ki

P

Q

ka ek single element hai.

1.9 Sets par Operations

Arithmetic mein ek operation do numbers leta hai aur ek naya number deta hai, jaise addition 7 aur 8 ko 15 bana deta hai. Set operations bhi aise hi kaam karte hain: woh do sets lete hain aur ek set dete hain. Ab se har set kisi universal set $U$ ka subset hai, aur Venn diagram mein har operation ka result ek shaded region hota hai.

1.9.1 Sets ka union

Maano school ka choir aur debate team dono ek hi bus se inter-school festival ja rahe hain. Bus ko har us student ko le jaana hai jo choir mein hai, debate team mein hai, ya dono mein hai, aur jo student dono mein hai use bhi sirf ek seat chahiye. Yeh mili-juli list dono sets ka union (सम्मिलन) hai.

Example 12NCERT Example 12 · p. 14

Union ki list banana

Maano $A = {3, 6, 9, 12, 15}$ aur $B = {5, 10, 15} .$ $A \cup B$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13NCERT Example 13 · p. 14

Subset ke saath union

Maano $A$ , 12 ke positive divisors ka set hai aur $B$ , 4 ke positive divisors ka set. Dikhao ki $A \cup B = A$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14NCERT Example 14 · p. 14

Bus mein kaun chadhega?

Class 11 ke chaar students school choir mein gaate hain, $C$ = {Anjali, Farhan, Gurpreet, Meera}, aur teen debate team mein hain, $D$ = {Farhan, Lakshmi, Meera}. $C \cup D$ nikalo aur batao ki yeh kya dikhata hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Definition

Union

Sets $A$ aur $B$ ka union, jise $A \cup B$ likhte hain aur " $A$ union $B$ " padhte hain, un elements ka set hai jo $A$ aur $B$ mein se kam se kam ek mein belong karte hain: $A \cup B = {x : x \in A or x \in B} .$

Example ${1, 2, 3} \cup {3, 4} = {1, 2, 3, 4};$ common element 3 ek hi baar likha jaata hai.

Yahan word "or" inclusive hai: jo element dono sets mein hai, woh bhi ginta hai. Venn diagram mein $A \cup B$ dono circles se dhaka poora region hai (Fig. 1.9).

3/3Sirf overlap shade karo: $A \cap B = {1, 2, 3, 6} .$

Fig. 1.9Union: kisi bhi circle mein jo kuch hai. Intersection: sirf overlap.

Rectangle U mein do overlap karte circles hain, left mein A aur right mein B. Overlap mein 1, 2, 3 aur 6 hain; A ke baaki hisse mein 4 aur 12; B ke baaki hisse mein 9 aur 18; numbers 5 aur 7 dono circles ke bahar hain. Pehle step mein kuch shaded nahi hai. Doosre step mein dono circles poore shade hote hain: union, aath numbers ke saath. Teesre step mein sirf overlap shade hota hai: intersection, chaar numbers 1, 2, 3 aur 6 ke saath.

Union paanch laws follow karta hai. Har law "or" ke matlab se nikalta hai, aur ek chhota sa Venn sketch har ek ko saaf kar deta hai.

Commutative law: $A \cup B = B \cup A,$ kyunki " $A$ mein ya $B$ mein" aur " $B$ mein ya $A$ mein" ek hi baat hai.
Associative law: $(A \cup B) \cup C = A \cup (B \cup C);$ dono woh elements hain jo teeno sets mein se kam se kam ek mein hain, isliye hum seedha $A \cup B \cup C$ likhte hain.
Identity law: $A \cup ϕ = A$ . Empty set kuch nahi jodta, isliye $ϕ$ , $\cup$ ki identity hai, jaise 0, $+$ ki identity hai.
Idempotent law: $A \cup A = A$ ; kisi set ko khud se jodne par koi naya element nahi aata.
$U$ ka law: $U \cup A = U$ , kyunki $A$ pehle se $U$ ke andar hai.

1.9.2 Sets ka intersection

Wapas festival par chalte hain. Organisers ko woh students bhi chahiye jo choir aur debate team dono mein hain, kyunki woh students ek saath do halls mein nahi ho sakte. Yeh common hissa intersection (सर्वनिष्ठ) hai.

Example 15NCERT Example 15 · p. 15

Ek common element

$A = {3, 6, 9, 12, 15}$ aur $B = {5, 10, 15},$ ke liye $A \cap B$ nikalo; yeh wahi do sets hain jo Example 12 mein the.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 16NCERT Example 16 · p. 15

Dono teams mein

Example 14 ke choir aur debate-team wale sets $C$ aur $D$ se $C \cap D$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 17NCERT Example 17 · p. 15

Subset ke saath intersection

Maano $A = {x : x \in 𝐍, x \leq 20}$ aur $B$ , $A$ mein perfect squares ka set hai. $A \cap B$ nikalo aur dikhao ki yeh $B$ ke equal hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Definition

Intersection

$A$ aur $B$ ka intersection, jise $A \cap B$ likhte hain aur " $A$ intersection $B$ " padhte hain, un elements ko ikattha karta hai jo $A$ aur $B$ mein common hain: $A \cap B = {x : x \in A and x \in B} .$

Example ${1, 2, 3} \cap {2, 3, 4} = {2, 3} .$

Venn diagram mein $A \cap B$ woh region hai jahan circles overlap karte hain (Fig. 1.9 ka aakhri step). Ho sakta hai do sets mein koi element common na ho, jaise Class 11 ke Section A aur Section B ke students, ya $𝐐$ aur $𝐓$ . Aise sets ko disjoint (असंयुक्त) kehte hain, aur unke circles alag-alag banaye jaate hain.

Definition

Disjoint sets

Sets $A$ aur $B$ disjoint hain agar unmein koi common element na ho, yaani agar $A \cap B = ϕ$ .

Example ${10, 20, 30}$ aur ${15, 25, 35}$ disjoint hain.

Intersection ke apne laws hain.

Commutative law: $A \cap B = B \cap A .$
Associative law: $(A \cap B) \cap C = A \cap (B \cap C);$ dono woh elements hain jo teeno sets mein common hain.
$ϕ$ aur $U$ ke laws: $ϕ \cap A = ϕ$ aur $U \cap A = A$ . Yahan $U$ identity ka kaam karta hai, jaise $\times$ ke liye 1 karta hai.
Idempotent law: $A \cap A = A$ .
Distributive law: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C);$ intersection, union par distribute hota hai.

Distributive law arithmetic ke $a (b + c) = a b + a c$ jaisa dikhta hai, aur Fig. 1.10 dono sides ko step by step banakar dikhata hai ki woh same region cover karti hain. Arithmetic ke ulat, operations ki jagah badalne par bhi yeh law sahi rehta hai: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C),$ yaani union bhi intersection par distribute hota hai, jabki zyaadatar numbers ke liye $a + b c \neq (a + b) (a + c)$ .

5/5Inka union, $(A \cap B) \cup (A \cap C),$ theek left side wala region cover karta hai.

Fig. 1.10Distributive law ki dono sides same region shade karti hain.

Rectangle U ke andar teen overlap karte circles A, B aur C. Pehle step mein kuch shaded nahi hai. Doosre step mein poora B aur C shade hota hai: set B union C. Teesre step mein sirf A ka woh hissa shade hota hai jo B ya C mein hai: ek curved region jo A ke B ke saath overlap aur A ke C ke saath overlap se bana hai. Chauthe step mein A aur B ka overlap ek taraf jhuki lines se hatch hota hai aur A aur C ka overlap doosri taraf jhuki lines se; beech ka chhota hissa, jahan teeno circles milte hain, dono hatchings rakhta hai. Paanchve step mein teesre step wali solid shading hatching ke neeche wapas aati hai aur theek hatched area ko cover karti hai.

InteractiveVenn lab

Rectangle ke andar do ya teen circles wala Venn diagram, ek keypad jismein letters A, B, C aur U, empty set, union, intersection, minus aur complement ke signs aur brackets hain, aur do expression boxes. Har expression ka region shade hota hai aur shabdon mein describe hota hai, saath mein U = {1, 2, …, 20} ke andar sample sets ke liye usmein aane wale elements bhi. Jab dono boxes mein expression ho, to ek panel batata hai ki kya woh sets ke har choice ke liye same regions cover karte hain, ya us region ka naam batata hai jahan woh alag hain. Challenge mode ek expression dikhata hai, tumse uske regions par tap karne ko kehta hai, aur tumhara answer check karta hai.

Fig. 1.11

A

B

aur

C

mein koi bhi expression type karo aur uska region dekho; kisi law ko test karne ke liye do expressions compare karo, ya khud ek region shade karke check karo.

1.9.3 Sets ka difference

Kuch questions poochte hain ki ek set mein kya hai jo doosre mein nahi. Choir ke kaun se members debate nahi karte? Kaun se primes odd nahi hain?

Definition

Sets ka difference

Difference $A - B$ , jise " $A$ minus $B$ " padhte hain, $A$ ke un elements ka set hai jo $B$ mein nahi hain: $A - B = {x : x \in A and x \notin B} .$

Example ${1, 2, 3, 4} - {3, 4, 5} = {1, 2} .$

Example 18NCERT Example 18 · p. 16

Woh primes jo odd nahi hain

Maano $A$ , 15 se chhote primes ka set hai aur $B$ , 15 se chhote odd natural numbers ka set. $A - B$ aur $B - A$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 19NCERT Example 19 · p. 17

Do shehron ke letters

Maano $X$ , INDORE ke letters ka set hai aur $Y$ , DELHI ke letters ka set. $X - Y$ aur $Y - X$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

To order se farak padta hai: general roop mein $A - B \neq B - A .$ Venn diagram mein $A - B$ , circle $A$ ka woh hissa hai jo circle $B$ ke bahar hai (Fig. 1.12).

Dono differences aur intersection $A \cup B$ ko teen tukdon mein kaatte hain. $A \cup B$ ka har element ya to sirf $A$ mein hai, ya sirf $B$ mein, ya dono mein, aur kabhi bhi inmein se do mein ek saath nahi. Isliye $A - B$ , $A \cap B$ aur $B - A$ pairwise disjoint hain, aur milkar $A \cup B$ banate hain.

4/4 $X - Y$ , $X \cap Y$ aur $Y - X$ kabhi overlap nahi karte, aur milkar $X \cup Y$ ko bharte hain.

Fig. 1.12Dono differences aur intersection, union ke teen aise tukde hain jo overlap nahi karte.

Ek rectangle mein overlap karte circles X aur Y hain. Overlap mein I, D aur E hain; X ke baaki hisse mein N, O aur R; Y ke baaki hisse mein L aur H; letters K aur M dono ke bahar hain. Pehle step mein kuch shaded nahi hai. Doosre step mein left crescent, yaani X minus Y, shade hota hai. Teesre step mein uski jagah right crescent, yaani Y minus X, shade hota hai. Chauthe step mein left crescent, overlap aur right crescent teen alag tints mein shade hote hain: teen alag tukde jo milkar dono circles ko bharte hain.

1.10 Set ka Complement

Ek school library Monday se Friday tak khuli rehti hai. Maano $U$ hafte ke saat dinon ka set hai aur $A$ un dinon ka set jab library khuli hai. Jin dinon library band rehti hai, woh theek $U$ ke woh din hain jo $A$ mein nahi hain: Saturday aur Sunday. Universal set ka yeh bacha hua hissa $A$ ka complement (पूरक) hai.

Definition

Complement

Universal set $U$ ke subset $A$ ke liye, complement $A^{'}$ mein $U$ ki har woh cheez aati hai jo $A$ mein nahi hai: $A^{'} = {x : x \in U, x \notin A},$ jo difference $U - A$ hi hai.

Example $U$ = hafte ke din aur $A$ = {Monday, …, Friday} ho, to $A^{'}$ = {Saturday, Sunday}.

Complement $U$ par depend karta hai. Agar $U$ sirf Monday se Saturday tak ke chhah din hota, to usi $A$ ka $A^{'}$ = {Saturday} hota. Venn diagram mein $A^{'}$ woh sab hai jo rectangle ke andar aur circle ke bahar hai (Fig. 1.13).

2/3 $A^{'}$ = {Sat, Sun}: $U$ ka woh hissa jo $A$ ke bahar hai.

Fig. 1.13Complement,

U

ka woh hissa hai jo set ke bahar hai.

Rectangle U mein hafte ke saat din hain. A label wale circle mein Mon, Tue, Wed, Thu aur Fri hain; Sat aur Sun uske bahar hain. Pehle step mein kuch shaded nahi hai. Doosre step mein rectangle ke andar lekin circle ke bahar sab kuch shade hota hai: A ka complement, jismein Sat aur Sun hain. Teesre step mein shading wapas circle ke andar aa jaati hai, jo dikhata hai ki complement ka complement phir se A hai.

Example 20NCERT Example 20 · p. 19

Complement nikalna

Maano $U = {1, 2, \dots, 12}$ aur $A$ , $U$ mein 3 ke multiples ka set hai. $A^{'}$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 21NCERT Example 21 · p. 19

Shabdon mein complement

Maano $U$ kisi school ke Class 11 students ka set hai aur $A$ un students ka set jo school bus se school aate hain. $A^{'}$ ko describe karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ab Example 20 wale $A^{'}$ ka complement lo: ${1, 2, 4, 5, 7, 8, 10, 11}$ ke bahar $U$ ke elements 3, 6, 9 aur 12 hain, jo phir se $A$ hai. Aisa hamesha hota hai. Bahar ke bahar ko hatane par andar wala wapas milta hai, isliye $U$ ke har subset $A$ ke liye $(A^{'})^{'} = A$ . Yeh bhi dhyaan do ki $A^{'}$ khud $U$ ka ek subset hai.

Example 22NCERT Example 22 · p. 19

Union ka complement

Maano $U = {a, b, c, d, e, f, g},$ $A = {a, b, c}$ aur $B = {c, d, e} .$ $A^{'}$ , $B^{'}$ , $A^{'} \cap B^{'}$ aur $(A \cup B)^{'},$ nikalo, aur aakhri do ko compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 22 koi ittefaq nahi hai. Koi element $A \cup B$ ke bahar theek tab hota hai jab woh $A$ ke bahar ho aur $B$ ke bhi bahar ho. Intersections ke liye bhi aisa hi ek statement sahi hai, aur is jodi ka naam British mathematician Augustus De Morgan (1806–1871) ke naam par hai, jinka janm Madurai mein hua tha.

LawDe Morgan's laws

$U$ ke kinhi bhi subsets $A$ aur $B$ ke liye, $A \cup B$ ka complement $A^{'} \cap B^{'}$ hai, aur $A \cap B$ ka complement $A^{'} \cup B^{'}$ hai. Complement lene se union aur intersection aapas mein badal jaate hain.

Hum pehla law $U$ ke kisi bhi element $x$ ko follow karke prove karte hain. $x \in (A \cup B)^{'}$ kehne ka matlab hai $x \notin A \cup B$ , yaani $x$ dono mein se kisi bhi set mein nahi hai: $x \notin A$ aur $x \notin B$ . Yeh wahi baat hai ki $x \in A^{'}$ aur $x \in B^{'}$ , jiska matlab hai $x \in A^{'} \cap B^{'} .$

Aage ki taraf padhne par yeh chain dikhati hai $(A \cup B)^{'} \subset A^{'} \cap B^{'};$ peeche ki taraf padhne par dikhati hai $A^{'} \cap B^{'} \subset (A \cup B)^{'} .$ Doosra law bhi isi tarah nikalta hai, kyunki $x \notin A \cap B$ ka matlab hai $x \notin A$ ya $x \notin B$ .

Venn diagram pehle law ko saaf dikha deta hai (Fig. 1.14): dono circles ke bahar ka hissa shade karne par wahi region milta hai jo $A$ ke bahar aur $B$ ke bahar ke hisson ko overlap karne par milta hai.

5/5 $(A \cup B)^{'}$ aur $A^{'} \cap B^{'}$ bilkul same region shade karte hain.

Fig. 1.14

(A \cup B)^{'}

aur

A^{'} \cap B^{'}

ek hi region hain.

Rectangle U mein do overlap karte circles A aur B. Pehle step mein dono circles shaded hain: union. Doosre step mein sirf dono circles ke bahar ka region shaded hai: union ka complement. Teesre step mein circle A ke bahar sab kuch ek taraf jhuki lines se hatch hota hai: A ka complement. Chauthe step mein circle B ke bahar sab kuch bhi doosri taraf jhuki lines se hatch hota hai; sirf dono circles ke bahar wale region par dono hatchings hain, jo dono complements ka intersection hai. Paanchve step mein doosre step wali solid shading halki hatching ke neeche wapas aati hai aur theek cross-hatched region ko bharti hai.

Complement chaar laws follow karta hai, aur har ek ko ek chhota Venn sketch confirm kar deta hai.

Complement laws: $A \cup A^{'} = U$ aur $A \cap A^{'} = ϕ;$ ek set aur uska complement bina overlap kiye $U$ ko bhar dete hain.
De Morgan's laws: $A \cup B$ ka complement $A^{'} \cap B^{'}$ hai, aur $A \cap B$ ka complement $A^{'} \cup B^{'}$ hai; complement lene se $\cup$ aur $\cap$ aapas mein badal jaate hain.
Double complement: $(A^{'})^{'} = A;$ bahar ka bahar, andar hi hai.
$ϕ$ aur $U$ ke laws: $ϕ^{'} = U$ aur $U^{'} = ϕ$ ; kuch bhi na chhodne par poora $U$ milta hai, aur $U$ ke bahar kuch nahi hai.

Practice · 7 questions

Practice 1.5: answers ke saath

1
Short answerEasyNCERT Exercise 1.5, Q1
Maano $U = {1, 2, \dots, 10},$ $A = {2, 4, 6, 8, 10},$ $B = {1, 2, 3, 4, 5}$ aur $C = {3, 6, 9} .$ Har set nikalo.
1. $A^{'}$
  Solution dikhao
  Answer ${1, 3, 5, 7, 9}$
  $U$ ke odd numbers.
2. $(A \cup C)^{'}$
  Solution dikhao
  Answer ${1, 5, 7}$
  $A \cup C = {2, 3, 4, 6, 8, 9, 10} .$
3. $(A \cap B)^{'}$
  Solution dikhao
  Answer ${1, 3, 5, 6, 7, 8, 9, 10}$
  $A \cap B = {2, 4} .$
4. $(B - C)^{'}$
  Solution dikhao
  Answer ${3, 6, 7, 8, 9, 10}$
  $B - C = {1, 2, 4, 5} .$
Short answerEasyNCERT Exercise 1.5, Q2
Maano $U = {p, q, r, s, t, u, v, w} .$ Inke complements nikalo: (a) ${p, q}$ (b) ${q, s, u, w}$ (c) ${r, s, t, u, v}$ (d) $U$ .
Solution dikhao
(a) ${r, s, t, u, v, w}$ (b) ${p, r, t, v}$ (c) ${p, q, w}$ (d) $ϕ$ , kyunki $U$ ka kuch bhi $U$ ke bahar nahi hai.
3
Short answerMediumNCERT Exercise 1.5, Q3
$𝐍$ ko universal set maankar har set ka complement likho.
1. 4 ke natural multiples
  Solution dikhao
  Answerwoh natural numbers jo 4 se divisible nahi hain: ${1, 2, 3, 5, 6, 7, 9, \dots}$
  $𝐍$ se 4, 8, 12, … hatao.
2. ${x : x \in 𝐍, x > 10}$
  Solution dikhao
  Answer ${1, 2, \dots, 10}$
  10 tak ke natural numbers bachte hain.
3. ${x : x \in 𝐍, 3 x - 2 = 13}$
  Solution dikhao
  Answer5 ke alawa saare natural numbers
  Set ${5}$ hai, kyunki $3 x = 15$ .
4. woh natural numbers jo 2 aur 3 dono se divisible hain
  Solution dikhao
  Answerwoh natural numbers jo 6 se divisible nahi hain
  2 aur 3 dono se divisible hone ka matlab hai 6 se divisible hona.
4
ProofMediumNCERT Exercise 1.5, Q4
Maano $U = {1, 2, \dots, 10},$ $A = {1, 3, 5, 7, 9}$ aur $B = {1, 2, 5, 10} .$ De Morgan's laws mein se har ek ko verify karo.
1. $(A \cup B)^{'} = A^{'} \cap B^{'}$
  Solution dikhao
  AnswerDono sides ka value ${4, 6, 8}$ hai.
  $A \cup B = {1, 2, 3, 5, 7, 9, 10},$ isliye $(A \cup B)^{'} = {4, 6, 8} .$
  $A^{'} = {2, 4, 6, 8, 10}$ aur $B^{'} = {3, 4, 6, 7, 8, 9},$ isliye $A^{'} \cap B^{'} = {4, 6, 8} .$
2. $(A \cap B)^{'} = A^{'} \cup B^{'}$
  Solution dikhao
  AnswerDono sides ka value hai ${2, 3, 4, 6, 7, 8, 9, 10} .$
  $A \cap B = {1, 5},$ isliye $(A \cap B)^{'}$ , $U$ mein baaki sab kuch hai.
  $A^{'} \cup B^{'} = {2, 4, 6, 8, 10} \cup {3, 4, 6, 7, 8, 9},$ jo wahi aath numbers hain.
5
DiagramMediumNCERT Exercise 1.5, Q5
Rectangle $U$ ke andar do overlap karte circles $A$ aur $B$ banao, aur har set ka region describe karo. Apne answers Venn lab mein check karo.
1. $A \cap B^{'}$
  Solution dikhao
  Answer $A$ ka woh crescent jo $B$ ke bahar hai, yaani wahi region jo $A - B$ hai.
  $A$ ke andar aur $B$ ke bahar.
2. $(A - B)^{'}$
  Solution dikhao
  AnswerUs crescent ke alawa sab kuch: poora $B$ aur saath mein dono circles ke bahar ka region.
  Crescent ka complement rectangle ka baaki hissa hai.
3. $A^{'} \cup B$
  Solution dikhao
  AnswerWahi region jo (b) mein hai.
  $A$ ke bahar, ya $B$ ke andar: isse $A$ ke us hisse ko chhodkar sab kuch cover ho jaata hai jo $B$ ke bahar hai. Sach mein, De Morgan's law se $(A - B)^{'} = (A \cap B^{'})^{'} = A^{'} \cup B$ .
4. $(A \cap B) \cup (A \cup B)^{'}$
  Solution dikhao
  AnswerCircles ka overlap aur saath mein dono ke bahar ka region.
  Yeh woh elements hain jo dono sets mein hain ya kisi mein bhi nahi.
Short answerMediumNCERT Exercise 1.5, Q6
Maano $U$ ek plane ke saare quadrilaterals ka set hai, aur $A$ un quadrilaterals ka set jinka kam se kam ek angle $90^{\circ}$ se alag hai. $A^{'}$ kya hai?
Solution dikhao
Answer $A^{'}$ rectangles ka set hai (squares samet).
Koi quadrilateral $A$ ke bahar theek tab hota hai jab uska koi bhi angle $90^{\circ}$ se alag na ho, yaani jab chaaron angles right angles hon. Aise quadrilaterals rectangles hain, aur squares khaas rectangles hain.
7
Quick answerEasyNCERT Exercise 1.5, Q7
$U$ ke subset $A$ ke baare mein true statements banane ke liye blanks bharo.
1. $U^{'} \cup A =$
  Answer dekho
  Answer A
  $U^{'} = ϕ$ , aur $ϕ \cup A = A$ .
2. $ϕ^{'} \cup A =$
  Answer dekho
  Answer U
  $ϕ^{'} = U$ , aur $U \cup A = U$ .
3. $(A^{'})^{'} \cap A^{'} =$
  Answer dekho
  Answer φ
  $(A^{'})^{'} = A,$ aur $A \cap A^{'} = ϕ .$
4. $A \cup (A^{'} \cap U) =$
  Answer dekho
  Answer U
  $A^{'} \cap U = A^{'},$ aur $A \cup A^{'} = U .$

Miscellaneous Examples

Example 23NCERT Example 23 · p. 20

Same letters, alag words

Words ASSAM aur MASS kaafi alag dikhte hain. Prove karo ki dono bilkul same letters se bane hain, yaani unke letters ke sets equal hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 24NCERT Example 24 · p. 21

Har subset ki list

Saare subsets list karo: $A = {x : x \in 𝐙, x^{3} = 4 x} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4/4 $2 \times 2 \times 2 = 8;$ isi tarah $n$ elements wale set ke $2^{n}$ subsets hote hain.

Fig. 1.15Teen haan/nahi choices se aath subsets milte hain. Solid branches ka matlab "lo", dashed branches ka matlab "chhodo".

Left se right badhne wala ek tree. Pehle step mein ek shuruaati dot heading "−2 lein?" ke neeche do branches mein batta hai: haan ke liye solid branch aur nahi ke liye dashed branch. Doosre step mein un dots mein se har ek "0 lein?" ke neeche phir se batta hai, aur chaar dots milte hain. Teesre step mein chaaron dots "2 lein?" ke neeche batte hain, aur aath end dots milte hain jinke labels upar se neeche hain {−2, 0, 2}, {−2, 0}, {−2, 2}, {−2}, {0, 2}, {0}, {2} aur khaali braces. Chauthe step mein tree ke neeche ginti aati hai: do guna do guna do barabar aath.

Example 25NCERT Example 25 · p. 21

Equal differences hon to sets bhi equal

Dikhao ki agar $A - B = B - A,$ to $A = B$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Agar

A \subset B

, to inmein se kya zaroor true hoga?

B - A = ϕ

A \cap B = B

A \cup B = A

A - B = ϕ

Answer dekho

Answer (D) $A - B = ϕ$

A

ka har element

B

mein hai, isliye

A - B

mein kuch nahi bachta. Baaki teen statements sirf tab sahi hain jab

A = B

Practice · 10 questions

Chapter 1 par miscellaneous exercises: answers ke saath

Short answerMediumNCERT Miscellaneous Exercise, Q1
Maano $A = {x : x \in 𝐑, x^{2} - 7 x + 10 = 0},$ $B = {2, 5, 8},$ $C$ un natural numbers ka set hai jo 3 se divide karne par remainder 2 dete hain, aur $D = {5}$ . Inmein se kaunse sets kin doosre sets ke subsets hain?
Solution dikhao
Answer $D \subset A \subset B \subset C,$ isliye $D \subset B$ , $D \subset C$ aur $A \subset C$ bhi.
$x^{2} - 7 x + 10 = (x - 2) (x - 5),$ isliye $A = {2, 5}$ , aur $C = {2, 5, 8, 11, \dots} .$
Tab $D \subset A$ (5 ek root hai), $A \subset B$ , aur $B \subset C$ kyunki 2, 5 aur 8 mein se har ek 3 se divide karne par remainder 2 deta hai. Subsets ke subsets bhi subsets hote hain, isse baaki teen milte hain. Koi aur inclusion sahi nahi hai: jaise $8 \in B$ lekin $8 \notin A$ .
2
ProofHardNCERT Miscellaneous Exercise, Q2
Decide karo ki har statement true hai ya false. True statements ko prove karo aur false ke liye counterexample do.
1. Jab bhi $A \subset B$ , har $x \in A$ , $x \in B$ ko bhi satisfy karta hai.
  SahiGalat
  Solution dikhao
  True: $A$ ka har element $B$ ka element hai, aur $x$ , $A$ ka element hai.
2. Jab bhi $B \in C$ aur $A \in B$ , to isse $A \in C$ nikalta hai.
  SahiGalat
  Solution dikhao
  False. Lo $A = {1}$ , $B = {{1}}$ aur $C = {{{1}}}$ . Tab $A \in B$ aur $B \in C$ , lekin $C$ ka akela element ${{1}} \neq A$ hai, isliye $A \notin C$ .
3. Jab bhi $B \subset C$ aur $A \subset̸ B$ , to isse $A \subset̸ C$ nikalta hai.
  SahiGalat
  Solution dikhao
  False. Lo $A = {1, 3}$ , $B = {1}$ , $C = {1, 3}$ : $A \subset̸ B$ (3 nahi hai) aur $B \subset C$ , phir bhi $A \subset C$ .
4. Jab bhi $B \subset̸ C$ aur $A \subset B$ , to isse $A \subset̸ C$ nikalta hai.
  SahiGalat
  Solution dikhao
  False. Lo $A = {1}$ , $B = {1, 2}$ , $C = {1}$ : $A \subset B$ aur $B \subset̸ C$ , phir bhi $A \subset C$ .
ProofHardNCERT Miscellaneous Exercise, Q3
Maano $A$ , $B$ aur $C$ aise sets hain ki $A - B = A - C$ aur $B - A = C - A .$ Dikhao ki $B = C$ .
Solution dikhao
Maano $x \in B$ . Agar $x \notin A$ , to $x \in B - A = C - A,$ isliye $x \in C$ .
Agar $x \in A$ , to $x \notin A - B$ (kyunki $x \in B$ ). Kyunki $A - B = A - C,$ $x \notin A - C$ bhi hai. $x \in A$ hai, isliye $x \in C$ hona hi chahiye.
To dono cases mein $B \subset C$ . Hypotheses $B$ aur $C$ mein symmetric hain, isliye $C \subset B$ bhi, aur $B = C$ .
ProofHardNCERT Miscellaneous Exercise, Q4
$U$ ke subsets $A$ aur $B$ ke liye dikhao ki yeh chaar conditions equivalent hain: (i) $A \cap B = ϕ$ , (ii) $A \subset B^{'}$ , (iii) $A - B = A$ , (iv) $B \subset A^{'}$ .
Solution dikhao
(i) ⇒ (ii): agar $x \in A$ , to $x \notin B$ (warna $x \in A \cap B$ hota), isliye $x \in B^{'}$ .
(ii) ⇒ (iii): $A - B \subset A$ hamesha. Agar $x \in A$ , to $x \in B^{'}$ , isliye $x \notin B$ aur $x \in A - B$ . To $A - B = A$ .
(iii) ⇒ (iv): maano $x \in B$ . Agar $x$ , $A$ mein hota, to woh $A - B$ mein hota, isliye $x \notin B$ , jo contradiction hai. To $x \in A^{'}$ .
(iv) ⇒ (i): agar $x \in A \cap B$ , to $x \in B \subset A^{'},$ isliye $x \notin A$ , jo contradiction hai. To $A \cap B = ϕ$ .
ProofMediumNCERT Miscellaneous Exercise, Q5
Agar $A \subset B$ , to dikhao ki $B^{'} \subset A^{'}$ .
Solution dikhao
Maano $x \in B^{'}$ , yaani $x \notin B$ . Agar $x$ , $A$ mein hota, to $x \in B$ hota kyunki $A \subset B$ , jo galat hai. To $x \notin A$ , yaani $x \in A^{'}$ . Isliye $B^{'} \subset A^{'}$ .
6
ProofMediumNCERT Miscellaneous Exercise, Q6
Dikhao ki $U$ ke kinhi bhi subsets $A$ aur $B$ ke liye, (a) $A - B = A \cap B^{'},$ aur (b) $A - (A - B) = A \cap B .$
1. $A - B = A \cap B^{'}$
  Solution dikhao
  $x \in A - B$ ka matlab hai $x \in A$ aur $x \notin B$ , jiska matlab hai $x \in A$ aur $x \in B^{'}$ , yaani $x \in A \cap B^{'} .$ Har step dono directions mein kaam karta hai, isliye sets equal hain.
2. $A - (A - B) = A \cap B$
  Solution dikhao
  (a) se, $A - (A - B) = A \cap (A \cap B^{'})^{'},$ aur De Morgan's law $(A \cap B^{'})^{'}$ ko $A^{'} \cup B$ mein badal deta hai.
  To set hai $A \cap (A^{'} \cup B) = (A \cap A^{'}) \cup (A \cap B) = ϕ \cup (A \cap B) = A \cap B .$
7
ProofMediumNCERT Miscellaneous Exercise, Q7
Sets ke laws use karke dikhao ki:
1. $(A \cup B) \cap (A \cup B^{'}) = A$
  Solution dikhao
  Union intersection par distribute hota hai, isliye left side hai $A \cup (B \cap B^{'}) = A \cup ϕ = A .$
2. $(A \cap B) \cup (A \cap B^{'}) = A$
  Solution dikhao
  Intersection union par distribute hota hai, isliye left side hai $A \cap (B \cup B^{'}) = A \cap U = A .$
Short answerMediumNCERT Miscellaneous Exercise, Q8
Ek example se dikhao ki $A \cup B = A \cup C$ hone se $B = C$ hona zaroori nahi.
Solution dikhao
AnswerJaise $A = {1, 2}$ , $B = {1}$ , $C = {2}$ .
$A = {1, 2}$ , $B = {1}$ aur $C = {2}$ lene par dono unions ${1, 2}$ ke equal hain, lekin $B \neq C$ . Jin elements mein $B$ aur $C$ alag hain, woh $A$ ke andar chhup jaate hain.
ProofHardNCERT Miscellaneous Exercise, Q9
Maano $A$ , $B$ aur $X$ , $U$ ke aise subsets hain ki $A \cup X = B \cup X$ aur $A \cup X^{'} = B \cup X^{'} .$ Dikhao ki $A = B$ .
Solution dikhao
Distributive law se, $(A \cup X) \cap (A \cup X^{'}) = A \cup (X \cap X^{'}) = A \cup ϕ = A .$ Isi tarah $(B \cup X) \cap (B \cup X^{'}) = B .$
Hypotheses se dono left-hand sides term by term equal hain, isliye $A = B$ .
Short answerMediumNCERT Miscellaneous Exercise, Q10
Aise sets $A$ , $B$ aur $C$ dhoondho, har ek mein teen elements, ki kinhi bhi do mein theek ek common element ho lekin $A \cap B \cap C = ϕ .$ $A \cup B \cup C$ mein kitne elements hain?
Solution dikhao
AnswerJaise $A = {1, 2, 3},$ $B = {3, 4, 5},$ $C = {5, 6, 1};$ union mein 6 elements hain.
$A \cap B = {3}$ , $B \cap C = {5}$ aur $A \cap C = {1}$ , jabki koi bhi number teeno sets mein nahi hai.
$A \cup B \cup C = {1, 2, 3, 4, 5, 6} :$ teen sets ki nau jagahon mein chhah alag numbers hain, kyunki 1, 3 aur 5 mein se har ek do baar gina gaya hai.

Recap

Summary aur key points

Set ek well-defined collection hai, jise roster form ya set-builder form mein describe kiya jaata hai (Section 1.2).
Empty set $ϕ$ mein koi element nahi hai; ${0}$ aur ${ϕ}$ empty nahi hain (Section 1.3).
$n (A)$ distinct elements ginta hai. Set finite hai agar woh empty ho ya uski ginti kisi natural number par khatam ho; warna woh infinite hai (Section 1.4).
Equal sets ke elements bilkul same hote hain; order aur repetition se kabhi farak nahi padta (Section 1.5).
$A \subset B$ jab $A$ ka har element $B$ mein ho. Hamesha $ϕ \subset A$ aur $A \subset A$ , aur $A = B$ theek tab jab $A \subset B$ aur $B \subset A$ (Section 1.6).
$\in$ ek element ko set se jodta hai, jabki $\subset$ do sets ko jodta hai (Example 11).
$𝐍 \subset 𝐙 \subset 𝐐 \subset 𝐑,$ aur irrationals $𝐓$ woh reals hain jo $𝐐$ ke bahar hain (Section 1.6.1).
End points $a < b$ wale interval ki length $b - a$ hoti hai, chahe woh open ho, closed ho ya half-open; square bracket apne end point ko include karta hai (Section 1.6.2).
Universal set $U$ mein discussion ke saare sets hote hain; Venn diagram ise rectangle ke roop mein aur iske subsets ko circles ke roop mein banata hai (Section 1.8).
$A \cup B$ mein woh hai jo kisi bhi set mein hai aur $A \cap B$ mein woh jo dono mein hai; disjoint sets ke liye $A \cap B = ϕ$ . Har operation doosre par distribute hota hai (Section 1.9.2).
$A - B$ mein $A$ ke woh elements rehte hain jo $B$ ke bahar hain. $A - B$ , $A \cap B$ aur $B - A$ milkar, bina overlap kiye, $A \cup B$ banate hain (Section 1.9.3).
$A^{'} = U - A$ aur $(A^{'})^{'} = A .$ De Morgan's laws se, $A \cup B$ ka complement $A^{'} \cap B^{'}$ hai, aur $A \cap B$ ka complement $A^{'} \cup B^{'}$ hai (Section 1.10).

Flash cards · 8

Ek nazar mein symbols

1 / 8

History seNCERT Historical Note · p. 22

Set theory kaise shuru hui

Georg Cantor (1845–1918) lagbhag 1870 mein sets tak pahunche, jab woh yeh sawaal pooch rahe the ki ek function ko sines aur cosines ke sum ke roop mein sirf ek hi tareeke se kab likha ja sakta hai. Answer kuch exceptional points ke ek set par depend karta tha, aur unhone aise sets ko khud unke liye study karna shuru kiya. Unki sabse mashhoor khoj 1874 mein aayi: real numbers ko 1, 2, 3, … numbered list mein nahi lagaya ja sakta, isliye infinite sets alag-alag sizes ke hote hain. Is idea ne mathematicians ko baant diya. Berlin mein Leopold Kronecker ne un proofs ko reject kiya jo kisi set ko banaye bina uske hone ka daava karte the, jabki Richard Dedekind (1831–1916) aur baad mein David Hilbert ne Cantor ka saath diya.

Lagbhag 1900 mein Gottlob Frege ne poori arithmetic ko logic aur sets par khada karne ki koshish ki. 1902 mein Bertrand Russell (1872–1970) ne unhe ek paradox likh bheja. Kisi set ko ordinary kaho agar woh khud ka element nahi hai, aur maano $R$ saare ordinary sets ka set hai. Agar $R$ ordinary hai, to woh $R$ mein belong karta hai, isliye woh ordinary nahi hai; agar woh ordinary nahi hai, to woh $R$ mein belong nahi karta, isliye woh ordinary hai. Dono tarah contradiction hai: koi bhi set sab kuch contain nahi kar sakta.

Iska ilaaj yeh tha ki kisi bhi description se set banana band kiya jaaye, aur uski jagah axioms ke roop mein un tareekon ki list banayi jaaye jinse purane sets se naye sets banaye ja sakte hain. Ernst Zermelo ne 1908 mein aise axioms diye; Abraham Fraenkel aur doosron ne inhe sudhaara, aur yeh ZF naam ka system ban gaya. Sets ko describe karne ka jo safe tareeka tum is chapter mein practise karte ho, yaani diye gaye universal set mein se elements chunna, theek wahi hai jiski in axioms mein ijaazat hai.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 1.1, Q1	Practice 1.1 · Q1
NCERT Exercise 1.1, Q2	Practice 1.1 · Q2
NCERT Exercise 1.1, Q3	Practice 1.1 · Q3
NCERT Exercise 1.1, Q4	Practice 1.1 · Q4
NCERT Exercise 1.1, Q5	Practice 1.1 · Q5
NCERT Exercise 1.1, Q6	Practice 1.1 · Q6
NCERT Exercise 1.2, Q1	Practice 1.2 · Q1
NCERT Exercise 1.2, Q2	Practice 1.2 · Q2
NCERT Exercise 1.2, Q3	Practice 1.2 · Q3
NCERT Exercise 1.2, Q4	Practice 1.2 · Q4
NCERT Exercise 1.2, Q5	Practice 1.2 · Q5
NCERT Exercise 1.2, Q6	Practice 1.2 · Q6
NCERT Exercise 1.3, Q1	Practice 1.3 · Q1
NCERT Exercise 1.3, Q2	Practice 1.3 · Q2
NCERT Exercise 1.3, Q3	Practice 1.3 · Q3
NCERT Exercise 1.3, Q4	Practice 1.3 · Q4
NCERT Exercise 1.3, Q5	Practice 1.3 · Q5
NCERT Exercise 1.3, Q6	Practice 1.3 · Q6
NCERT Exercise 1.3, Q7	Practice 1.3 · Q7
NCERT Exercise 1.3, Q8	Practice 1.3 · Q8
NCERT Exercise 1.4, Q1	Practice 1.4 · Q1
NCERT Exercise 1.4, Q2	Practice 1.4 · Q2
NCERT Exercise 1.4, Q3	Practice 1.4 · Q3
NCERT Exercise 1.4, Q4	Practice 1.4 · Q4
NCERT Exercise 1.4, Q5	Practice 1.4 · Q5
NCERT Exercise 1.4, Q6	Practice 1.4 · Q6
NCERT Exercise 1.4, Q7	Practice 1.4 · Q7
NCERT Exercise 1.4, Q8	Practice 1.4 · Q8
NCERT Exercise 1.4, Q9	Practice 1.4 · Q9
NCERT Exercise 1.4, Q10	Practice 1.4 · Q10
NCERT Exercise 1.4, Q11	Practice 1.4 · Q11
NCERT Exercise 1.4, Q12	Practice 1.4 · Q12
NCERT Exercise 1.5, Q1	Practice 1.5 · Q1
NCERT Exercise 1.5, Q2	Practice 1.5 · Q2
NCERT Exercise 1.5, Q3	Practice 1.5 · Q3
NCERT Exercise 1.5, Q4	Practice 1.5 · Q4
NCERT Exercise 1.5, Q5	Practice 1.5 · Q5
NCERT Exercise 1.5, Q6	Practice 1.5 · Q6
NCERT Exercise 1.5, Q7	Practice 1.5 · Q7
NCERT Miscellaneous Exercise, Q1	Chapter 1 par miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Chapter 1 par miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Chapter 1 par miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Chapter 1 par miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Chapter 1 par miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Chapter 1 par miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7	Chapter 1 par miscellaneous exercises · Q7
NCERT Miscellaneous Exercise, Q8	Chapter 1 par miscellaneous exercises · Q8
NCERT Miscellaneous Exercise, Q9	Chapter 1 par miscellaneous exercises · Q9
NCERT Miscellaneous Exercise, Q10	Chapter 1 par miscellaneous exercises · Q10

Quick check: MCQs

1.2 Sets aur unke Representations4 questions
1.3 Empty Set2 questions
1.4 Finite aur Infinite Sets2 questions
1.5 Equal Sets2 questions
1.6.2 R ke subsets ke roop mein intervals4 questions
1.8 Venn Diagrams2 questions
1.9.3 Sets ka difference4 questions
1.10 Set ka Complement4 questions
Miscellaneous Examples2 questions

Glossary · 16

Key terms (English + Hindi)

complementपूरकpoorak: $A^{'}$ , universal set $U$ ke un elements ka set jo $A$ mein nahi hain; isliye $A^{'} = U - A .$ Chapter mein pehli baar yahan aaya →

difference of setsसमुच्चयों का अंतरsamuchchayon ka antar: $A - B$ , $A$ ke un elements ka set jo $B$ mein nahi hain.
disjoint setsअसंयुक्त समुच्चयasanyukt samuchchay: Aise sets jinmein koi common element nahi, yaani $A \cap B = ϕ$ .Chapter mein pehli baar yahan aaya →

empty setरिक्त समुच्चयrikt samuchchay: Woh set jismein koi element nahi, jise $ϕ$ ya ${}$ likhte hain.
equal setsसमान समुच्चयsamaan samuchchay: Bilkul same elements wale sets; listing mein order aur repetition se farak nahi padta.

finite setपरिमित समुच्चयparimit samuchchay: Aisa set jo empty ho ya jiske elements gine ja sakein aur ginti kisi natural number par khatam ho; baaki har set infinite hai.

intersectionसर्वनिष्ठsarvanishth: $A \cap B$ , un elements ka set jo $A$ aur $B$ dono mein belong karte hain.Chapter mein pehli baar yahan aaya →
intervalअंतरालantaraal: Do end points ke beech ke saare real numbers ka set, jismein har end point included ya excluded hota hai, jaise $[a, b)$ .

proper subsetउचित उपसमुच्चयuchit upasamuchchay: $B$ ka aisa subset $A$ jismein $A \neq B$ , yaani $B$ mein kam se kam ek element $A$ ke bahar hai.Chapter mein pehli baar yahan aaya →

roster formरोस्टर रूपroster roop: Set ka aisa description jo uske elements ko braces ke andar, commas se alag karke list karta hai, jaise ${2, 4, 6}$ .Chapter mein pehli baar yahan aaya →

setसमुच्चयsamuchchay: Objects ka well-defined collection, jinhe uske elements kehte hain: kisi bhi object ke liye pakka pata hota hai ki woh belong karta hai ya nahi.Chapter mein pehli baar yahan aaya →
set-builder formसमुच्चय निर्माण रूपsamuchchay nirmaan roop: Set ka aisa description jo ek property se kiya jaata hai jo uske saare elements mein hoti hai, jaise ${x : x \in 𝐍, x < 4} .$ Chapter mein pehli baar yahan aaya →
subsetउपसमुच्चयupasamuchchay: $A \subset B$ jab $A$ ka har member $B$ mein bhi belong karta ho.

unionसम्मिलनsammilan: $A \cup B$ , un elements ka set jo $A$ mein ya $B$ mein ya dono mein hain.Chapter mein pehli baar yahan aaya →
universal setसार्वत्रिक समुच्चयsaarvatrik samuchchay: Woh set $U$ jismein kisi context mein discuss ho rahe saare sets aa jaate hain.

Venn diagramवेन आरेखven aarekh: Sets ki ek picture: universal set ke liye rectangle aur uske subsets ke liye closed curves, aam taur par circles.Chapter mein pehli baar yahan aaya →

Sets: Class 11 Maths Chapter 1 Hinglish mein

Is chapter mein aap kya seekhenge

1.1 Introduction

1.2 Sets aur unke Representations

Practice 1.1: answers ke saath

1.3 Empty Set

1.4 Finite aur Infinite Sets

1.5 Equal Sets

Practice 1.2: answers ke saath

1.6 Subsets

1.6.1 Real numbers ke set ke subsets

1.6.2 R ke subsets ke roop mein intervals

1.7 Universal Set

Practice 1.3: answers ke saath

1.8 Venn Diagrams

1.9 Sets par Operations

1.9.1 Sets ka union

1.9.2 Sets ka intersection

1.9.3 Sets ka difference

Practice 1.4: answers ke saath

1.10 Set ka Complement

Practice 1.5: answers ke saath

Miscellaneous Examples

Chapter 1 par miscellaneous exercises: answers ke saath

Practice questions answers ke saath (NCERT “Entrance jaisi practice”)

Summary aur key points

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye